The above question is based on this answer to a similar question. I just want to apply what has been pointed out in that answer to this question.

So we are interested in number of homomorphisms $f:A_5 \to S_5$ (if any). Now following the ideas given in the linked answer, $A_5$ is simple, so it has to normal subgroups, viz $\{e\}$ and $A_5$.

If $f$ is a homomorphism then ker$f$ is a normal subgroup of $A_5$.

  1. If ker$f$=$\{e\}$ , then $A_5/\{e\}\sim f(A_5)$, and so $f(A_5)$ is a subgroup of $S_5$ of order $60$.
  2. If ker$f=A_5$, then $f$ is trivial.

Also we know that "Any homomorphism from a simple group to any group is either trivial or injective."

Now in the view of above argument what can I conclude? It seems to me that only I can conclude that any homomorphism is injective. But how to calculate how many are there? Is it a valid approach or do we need a new approach?


  • $\begingroup$ Also we know that "Any homomorphism from a simple group to any group is either trivial or injective." $-$ You have shown that already in $(1)$ and $(2).$ $\endgroup$
    – little o
    Mar 4 '19 at 7:38
  • $\begingroup$ I think $A_5$ is the only subgroup of $S_5$ of order $60.$ So in order to find the number of injective homomorphisms from $A_5$ to $S_5$ you need only to find the number of elements in $\text {Aut} (A_5) = \left |A_5/Z(A_5) \right| = |A_5| = 60,$ since $Z(A_5) = \{e \},$ where $Z(A_5)$ is the center of $A_5.$ $\endgroup$
    – little o
    Mar 4 '19 at 7:48
  • $\begingroup$ For the proof of the above fact see here math.stackexchange.com/q/822096/543867 $\endgroup$
    – little o
    Mar 4 '19 at 7:59
  • $\begingroup$ "If ker$f$=$\{e\}$ , then $A_5/\{e\}\sim f(A_5)$" What does "$\sim$" mean in this context? $\endgroup$ Mar 4 '19 at 8:15

As you observed, the non-constant homomorphisms are injective. Because $A_5$ is the only subgroup of $S_5$ of order $60$ we are looking for automorphisms from $A_5$ to itself.

To each element $g\in S_5$ we get a conjugation automorphism $\phi_g(x)=gxg^{-1}$ for all $x\in A_5$. Because the centralizer of $A_5$ in $S_5$ is trivial, distinct choices of $g$ yield distinct automorphisms $\phi_g$.

Claim. Any automorphism $\phi$ of $A_5$ is of the form $\phi_g$ for some $g\in S_5$.

Proof. The group $A_5$ has five distinct Sylow $2$-subgroups. Namely $$P_5=\{1,(12)(34),(13)(24),(14)(23)\}$$ and its conjugates, each stabilizing a single element of $J_5:=\{1,2,3,4,5\}$. I will denote by $P_i$ the conjugate stabilizing $i\in J_5$. Because $\phi$ is an automorphism it must permute these 5 groups. So there is a permutation $\sigma\in S_5$ such that $\phi(P_i)=P_{\sigma(i)}$ for all $i\in J_5$.

On the other hand, the conjugation $\phi_\sigma$ also maps $P_i$ to $P_{\sigma(i)}$. Therefore the automorphism $\tau:=\phi\circ\phi_{\sigma^{-1}}$ has the property that $\tau(P_i)=P_i$ for all $i\in J_5$. Consequently also the normalizers are preserved: $\tau(N_{A_5}(P_i)=N_{A_5}(P_i)$ for all $i\in J_5$. But those normalizers are conjugates of $A_4$, each the stabilizer of an element of $J_5$. Consider a 3-cycle, such as $\alpha=(234)$. The only two 3-cycles normalizing both $P_1$ and $P_5$ are $\alpha$ and $\alpha^{-1}=(243)$. Therefore we must have $\tau(\alpha)=\alpha$ or $\tau(\alpha)=\alpha^{-1}$. But, we have $$\tau(\alpha P_2\alpha^{-1})=\tau(P_3)=P_3$$ as well as $$\tau(\alpha P_2\alpha^{-1})=\tau(\alpha)P_2\tau(\alpha)^{-1}=P_{\tau(\alpha)(2)}.$$ So we must have $\tau(\alpha)(2)=3$, leaving $\tau(\alpha)=\alpha$ as the only possibility.

It follows that $\tau(\beta)=\beta$ for all 3-cycles $\beta\in A_5$. But the 3-cycles generate $A_5$, so $\tau$ must be the identity mapping. Therefore $\phi=\phi_{\sigma}$. QED.

It follows that there are 120 injective homomorphisms from $A_5$ to $S_5$ and the trivial constant homomorphism.

  • $\begingroup$ What was going wrong in my argument in the comments above @Jyrki Lahtonen? $\endgroup$
    – little o
    Mar 4 '19 at 8:09
  • 1
    $\begingroup$ The same argument goes thru for $A_n$, $n\ge7$. The case $n=6$ is a famous exception for $S_6$ when the inner automorphisms form an index two subgroup of the automorphism group. See here for $n\ge7$. $\endgroup$ Mar 4 '19 at 8:09
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    $\begingroup$ Oh! Sorry. I have calculated order of $\text {Inn} (G).$ Right? $\endgroup$
    – little o
    Mar 4 '19 at 8:12
  • 1
    $\begingroup$ @Dbchatto67 I am not aware of one. I suspect it is difficult in general. Let's wait for an expert. $\endgroup$ Mar 4 '19 at 8:16
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    $\begingroup$ Possibly @KushalBhuyan. If you can prove some other way that the obvious conjugates of $A_4$ are the only subgroups isomorphic to $A_4$, and hence $\phi$ must permute them. The reason why I used Sylow 2-subgroups is that, as a set, they are characteristic to the group $A_5$, and hence any automorphism must permute them. $\endgroup$ Mar 4 '19 at 11:14

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